Generating Functions in Discrete Math a)Find the coefficient of $x^3y^4$ in $(2x + 5y)^7$. b) Find the coefficient of $x^5$ in $(3x -1)(2x +1)^8$.
I know this has to do with generating functions , but i'm not sure how to start with this problem in order to find the coefficient of part a) and b)
 A: You can (and should) do this with the binomial theorem. We expand $(2x+5y)^7$ this way:
$$\binom{7}{0}(2x)^7(5y)^{0} + \binom{7}{1}(2x)^6(5y)^{1} + \binom{7}{2}(2x)^5(5y)^{4} + \dots + \binom{7}{7}(2x)^0(5y)^{7} $$
Specifically, the term we are interested in is 
$$\binom{7}{4}(2x)^3(5y)^4$$
which can be simplified to $175000 \cdot x^3y^4$.
The second problem is done in the same way.
A: For (b), you have $(3x - 1)(2x + 1)^{8}$. So consider $(2x+1)^{8}$. You are interested in the coefficients of $x^{4}$ and $x^{5}$ in that term. When you multiply by $(3x-1)$ you can form $x^{5}$ by $3x * kx^{4}$, and $x^{5}$ is held constant by multiplying with $-1$. So $(2x + 1)^{8}$ has coefficient of $x^{5}$ as $\binom{8}{5} (2x)^{5}$ and coefficient of $x^{4}$ as $\binom{8}{4}$.
Now consider $3 * \binom{8}{4} x^{4} * x - \binom{8}{5} (2x)^{5}$.
A: A related technique. Recalling the Taylor series in two variables the coefficient of $x^3y^4$ is given by

$$\frac{1}{3!4!} \frac{\partial^3}{\partial x^3}\frac{\partial^4}{\partial y^4}(2x+5y)^7\Big |_{(x,y)=(0,0)} = 175000.  $$

A: Write(b) as:
\begin{align}
[x^5] (3 x - 1) (2 x + 1)^8
  &= 3 [x^4] (2 x + 1)^8 - [x^5](2 x + 1)^8 \\
  &= 3 \binom{8}{4} \cdot 2^4 - \binom{8}{5} \cdot 2^5
\end{align}
Use of the "coefficient of" ($[\cdot]$) operator simplifies writing and manipulating such expressions a lot.
